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Chapter 2: Problem 31

The process by which we determine limits of rational functions applies equally well to ratios containing non integer or negative powers of \(x\) :Divide numerator and denominator by the highest power of \(x\) in the denominator and proceed from there. $$\lim _{x \rightarrow \infty} \frac{2 x^{5 / 3}-x^{1 / 3}+7}{x^{8 / 5}+3 x+\sqrt{x}}$$

Short Answer

Expert verified
The limit is \(\infty\).

Step by step solution

01

Identify the Highest Power in the Denominator

To simplify the expression, we first need to determine the highest power of \(x\) in the denominator. The terms in the denominator are \(x^{8/5}\), \(3x\), and \(\sqrt{x}\), which is \(x^{1/2}\) in fractional form. The highest power of \(x\) among these terms is \(x^{8/5}\).
02

Divide Every Term by the Highest Power

Divide each term in both the numerator and the denominator by \(x^{8/5}\), the highest power from the denominator:\[\frac{2x^{5/3} - x^{1/3} + 7}{x^{8/5} + 3x + \sqrt{x}} = \frac{\frac{2x^{5/3}}{x^{8/5}} - \frac{x^{1/3}}{x^{8/5}} + \frac{7}{x^{8/5}}}{\frac{x^{8/5}}{x^{8/5}} + \frac{3x}{x^{8/5}} + \frac{\sqrt{x}}{x^{8/5}}}\]
03

Simplify the Expression

Simplify the fractions by subtracting the powers of \(x\) in each term:- Numerator: \(\frac{2x^{5/3}}{x^{8/5}} = 2x^{5/3 - 8/5} = 2x^{1/15}\), \(\frac{x^{1/3}}{x^{8/5}} = x^{1/3 - 8/5} = x^{-13/15}\), and \(\frac{7}{x^{8/5}} = 7x^{-8/5}\).- Denominator: \(\frac{x^{8/5}}{x^{8/5}} = 1\), \(\frac{3x}{x^{8/5}} = 3x^{-3/5}\), \(\frac{\sqrt{x}}{x^{8/5}} = x^{1/2 - 8/5} = x^{-3/10}\).
04

Evaluate the Limit

As \(x\) approaches infinity, terms with negative exponents \(x^{-a}\) where \(a > 0\) tend towards zero. Thus:- The numerator becomes \(2x^{1/15} + 0 + 0\), and since \(2x^{1/15}\) grows without bounds, it simplifies to infinity.- The denominator becomes \(1 + 0 + 0\), simplifying to 1.Therefore the limit is \(\lim_{x \to \infty} \frac{2x^{1/15}}{1} = \infty\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Non-integer Powers
When dealing with limits of rational functions, sometimes you'll encounter expressions with non-integer powers, like those in our exercise: \(x^{5/3}\), \(x^{8/5}\), \(x^{1/3}\). These are powers where the exponent is not a whole number, for example, a fraction. Non-integer powers can represent roots or fractional exponents, making calculations potentially tricky. For instance, \(x^{1/3}\) is the cube root of \(x\), and \(x^{5/3}\) is \(x\) raised to the power of \(5\) and then taking the cube root.
  • Expressions with non-integer powers can simplify in unique ways compared to integer exponents.
  • When working through limits, these powers are often broken down using properties of exponents to simplify the expression further.
Understanding how to manipulate these powers is crucial for simplifying expressions and evaluating limits.
Negative Powers
Negative powers often appear when simplifying rational functions by dividing every term by the highest power of \(x\). A negative power, such as \(x^{-a}\), indicates that \(x\) is in the denominator, like \(\frac{1}{x^a}\). In our exercise, when we divided by the highest power from the denominator, some terms in the numerator and denominator transformed into negative powers.
  • Example: \(x^{-13/15}\) translates to \(\frac{1}{x^{13/15}}\).
  • Negative powers are essential because as \(x\) approaches infinity, any term with a negative power tends towards zero.
This limit behavior simplifies computation because terms that shrink to zero can often be eliminated from the limit evaluation.
Infinity
Limits involving infinity deal with what happens to a function as \(x\) approaches extremely large values, either positive or negative. In our case, we evaluate the function's behavior as \(x\) approaches positive infinity. Observing how each term behaves for growing \(x\) values is key.
  • Terms with positive powers of \(x\), such as \(2x^{1/15}\), increase infinitely since their values grow without bound.
  • Negative power terms shrink towards zero.
When the highest power (or those powers close to zero) dominates in the numerator, while other terms diminish, the overall limit tends to highlight those growing parts leading the function to infinity.
Simplification Techniques
Simplification techniques focus on breaking down complex expressions to make them more manageable, particularly when calculating limits as \(x\) approaches infinity. In this exercise, we divide every term by the highest power in the denominator, allowing us to simplify each component.
  • This technique removes the complexity by converting higher powers into simpler forms, often resulting in unit fractions or removal of terms that approach zero.
  • By reducing the expression to its fundamental parts, we can more easily understand how each component of the function behaves as \(x\) increases.
This strategy is vital in limit calculations, ensuring that the expression is in its simplest form to evaluate the limit appropriately.

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Most popular questions from this chapter

To prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations. $$x(x-1)^{2}=1 \quad \text { (one root) }$$

To prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations. $$x^{x}=2$$

To prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations. $$\sqrt{x}+\sqrt{1+x}=4$$

To prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations. \(x^{3}-15 x+1=0 \quad\) (three roots)

Give an example of a function \(f(x)\) that is continuous for all values of \(x\) except \(x=2\). where it has a removable discontinuity. Explain how you know that \(f\) is discontinuous at \(x=2,\) and how you know the discontinuity is removable.
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